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      <title>Java死锁不是会让CPU爆表吗 - 交互式测验</title>
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            <div class="quiz-header">
              <h1>Java死锁不是会让CPU爆表吗 - 交互式测验</h1>
              <p>检验你对文档内容的理解程度</p>
            </div>
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  {
    "question": "根据文档描述，当多个线程发生纯粹的死锁并进入BLOCKED状态时，为什么CPU使用率通常不会直接飙升？",
    "options": [
      "因为BLOCKED状态的线程仍在后台执行低优先级的计算任务。",
      "因为操作系统会自动终止处于BLOCKED状态的线程以释放CPU资源。",
      "因为进入BLOCKED状态的线程会主动让出CPU时间片，不再参与CPU调度。",
      "因为JVM的垃圾回收器（GC）会优先处理BLOCKED线程，从而降低了CPU负载。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "文档中明确指出，当线程因等待监视器锁（synchronized）而进入BLOCKED状态时，它会主动放弃CPU时间片，不会消耗CPU。因此，纯粹的死锁本身并不会直接导致CPU使用率升高，就像堵车时司机熄火等待一样。",
    "source": "文档中「二、死锁到底会不会拉高CPU？」和「三、线程的六种状态」部分",
    "hint": "可以回顾一下文档中关于『BLOCKED』状态的描述及其对CPU占用的影响。"
  },
  {
    "question": "文档中提到了一个真实案例：某支付系统发生死锁后，CPU使用率飙升。以下哪项是导致CPU飙升的“间接原因”之一？",
    "options": [
      "大量线程进入TIMED_WAITING状态，不断唤醒消耗CPU。",
      "死锁导致JVM内存泄漏，触发了频繁的Full GC。",
      "线程池因大量工作线程被阻塞，疯狂创建新线程来补偿，导致资源消耗激增。",
      "操作系统内核检测到死锁后，启动了高耗能的诊断程序。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "文档中明确列出了CPU在死锁场景下升高的三大间接原因，其中“线程池的报复性补偿”是核心原因之一。当工作线程被锁阻塞时，线程池会误以为线程不足，从而创建大量新线程，这些新线程的创建和调度过程会消耗大量CPU资源。",
    "source": "文档中「二、死锁到底会不会拉高CPU？」部分的“CPU升高的三大间接原因”",
    "hint": "思考一下，当线程池里的员工（线程）都『堵在路上』时，线程池的管理者会怎么做？"
  },
  {
    "question": "文档中用“四辆车在十字路口互不相让”的比喻来解释死锁的哪个必要条件？",
    "options": [
      "互斥条件 (Mutual Exclusion)",
      "请求与保持 (Hold and Wait)",
      "不可剥夺 (No Preemption)",
      "循环等待 (Circular Wait)"
    ],
    "correctAnswerIndex": 3,
    "explanation": "“循环等待”指的是多个进程或线程形成一个环形的资源等待链，每个实体都在等待下一个实体所持有的资源。文档中用四辆车在路口互相挡住对方去路的比喻，生动地描绘了这种环形依赖关系，这是形成死锁的最后一个必要条件。",
    "source": "文档中「一、为什么线程会“卡死”？」部分的“死锁的四个必要条件”",
    "hint": "这个比喻形象地描述了一个『你等我，我等他，他又等你』的闭环。"
  },
  {
    "question": "为了预防在转账操作中发生死锁，文档中推荐的“锁排序法”是如何破坏死锁条件的？",
    "options": [
      "它通过确保所有线程总是按相同的顺序获取锁，从而破坏了“循环等待”条件。",
      "它通过允许高优先级的线程抢占低优先级线程的锁，破坏了“不可剥夺”条件。",
      "它将多个锁合并为一个全局锁，从而破坏了“请求与保持”条件。",
      "它使用了非独占锁，允许多个线程同时访问资源，破坏了“互斥条件”。"
    ],
    "correctAnswerIndex": 0,
    "explanation": "锁排序法通过一个固定的规则（例如，根据账户ID的大小）来决定获取锁的顺序。这样一来，所有线程都将遵循相同的加锁路径，无法形成A等B、B等A的环形等待链，从而有效地破坏了“循环等待”这一死锁的必要条件。",
    "source": "文档中「五、预防死锁的六个狠招」部分的“1. 锁排序法”",
    "hint": "思考一下，如果所有车辆都遵守『先让右侧车辆』的规则，十字路口的僵局还会发生吗？"
  },
  {
    "question": "根据「线程的六种状态」表格，一个线程调用了 `Thread.sleep(1000)` 方法后，它会进入哪种状态，并且这种状态是否消耗CPU？",
    "options": [
      "RUNNABLE，高CPU占用",
      "WAITING，无CPU占用",
      "TIMED_WAITING，无CPU占用",
      "BLOCKED，无CPU占用"
    ],
    "correctAnswerIndex": 2,
    "explanation": "文档中的表格明确指出，`Thread.sleep()` 这样的有时限等待操作会使线程进入TIMED_WAITING状态。在这种状态下，线程会暂时让出CPU，直到设定的时间结束或被中断，因此它不消耗CPU资源。",
    "source": "文档中「三、线程的六种状态」部分的表格",
    "hint": "这个状态的名字里就包含了『时间』的含义，并且它是一种『等待』状态。"
  },
  {
    "question": "在「一个真实案例全流程复盘」中，排查线上CPU飙升问题的第一个命令行操作是什么？",
    "options": [
      "使用 `jstack <pid>` 生成完整的线程堆栈。",
      "使用 `grep -A 20 \"deadlock\"` 在日志中搜索死锁。",
      "使用 `top -Hp <pid>` 找出消耗CPU最高的线程。",
      "使用 `printf \"%x\" <tid>` 将线程ID转换为十六进制。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "根据文档复盘的排查过程，第一步是使用 `top -Hp <pid>` 命令。这个命令可以按线程展示进程的资源使用情况，从而快速定位到具体是哪个线程导致了CPU占用率过高，为后续使用jstack精确分析该线程的行为奠定了基础。",
    "source": "文档中「六、一个真实案例全流程复盘」部分的“排查过程”",
    "hint": "在解决CPU高的问题时，首要任务是找到『罪魁祸首』是哪个具体的线程。"
  },
  {
    "question": "文档中提到的`tryLock`机制相比于`synchronized`块，在避免死锁方面提供了什么关键优势？",
    "options": [
      "`tryLock`的执行效率远高于`synchronized`。",
      "`tryLock`可以自动检测并断开循环等待链。",
      "`tryLock`允许线程在指定时间内尝试获取锁，如果失败则可以放弃或重试，避免了无限期阻塞。",
      "`tryLock`使用了更少的内存，适合资源受限的环境。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "`tryLock`的核心优势在于它提供了非阻塞或限时阻塞的加锁方式。线程可以尝试获取锁，如果锁被占用，它可以立即返回或等待一个设定的超时时间。这种机制打破了“请求与保持”的僵局，因为线程不必一直持有第一个锁去无限期地等待第二个锁，从而有效避免了死锁的发生。",
    "source": "文档中「五、预防死锁的六个狠招」部分的“2. 使用tryLock”",
    "hint": "想象一下打电话，如果占线，你是会一直举着电话等，还是挂掉待会儿再打？`tryLock`提供了后一种选择。"
  }
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                '<div class="score-card">' +
                '<h3>正确率</h3>' +
                '<p>' + accuracy + '%</p>' +
                '</div>' +
                '<div class="score-card">' +
                '<h3>错误率</h3>' +
                '<p>' + errorRate + '%</p>' +
                '</div>' +
                '<div class="score-card score-details">' +
                '<table>' +
                '<tr><td>正确</td><td>' + correctCount + '</td></tr>' +
                '<tr><td>错误</td><td>' + wrongCount + '</td></tr>' +
                '<tr><td>未答</td><td>' + skippedCount + '</td></tr>' +
                '</table>' +
                '</div>' +
                '</div>' +
                '<div class="report-actions">' +
                '<button id="review-btn">回顾测验</button>' +
                '<button class="primary" onclick="location.reload()">再试一次</button>' +
                '</div>' +
                '</div>';

            let reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }
            quizContentEl.insertAdjacentHTML('beforeend', finalScoreHtml);
            document.getElementById('review-btn').addEventListener('click', reviewQuiz);
        }

        function reviewQuiz() {
            const reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }

            document.querySelector('.quiz-body').style.display = 'block';
            document.getElementById('quiz-footer').style.display = 'flex';
            
            currentQuestionIndex = 0;
            loadQuestion();
            updateNavigationButtons(); // Ensure nav buttons are correctly displayed
        }

        // Event Delegation for options
        optionsListEl.addEventListener('click', (event) => {
            if (event.target && event.target.matches('li.option-item')) {
                // Check if the question has already been answered
                if (userAnswers[currentQuestionIndex] === null) {
                    const selectedIndex = parseInt(event.target.dataset.index, 10);
                    selectOption(selectedIndex);
                }
            }
        });

        prevBtn.addEventListener('click', goToPreviousQuestion);
        nextBtn.addEventListener('click', goToNextQuestion);
        hintBtn.addEventListener('click', showHint);
        
        // 初始加载
        loadQuestion();
      </script>
    </body>
    </html>
  